1/2*3+1/3*4+1/4*5+1/5*6+1/6*7+...+1/98*99+1/99*100
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{50}{100}-\frac{1}{100}\)
\(=\frac{49}{100}\)
1/2 x 3 + 1/3 x 4 + 1/4 x 5 + 1/5 x 6 + 1/6 x 7 + ....... + 1/98 x 99 + 1/99 x 100
= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ..... + 1/98 - 1/99 + 1/99 - 1/100
= 1/2 - 1/100
= 49/100
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
ta có \(\frac{1}{n\left(n+1\right)}=\frac{\left(n+1\right)-n}{n\left(n+1\right)}\)
\(=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}\)
\(=\frac{1}{n}-\frac{1}{n+1}\)
thay vào biểu thức ban đầu được
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
=\(=\frac{49}{100}\)
1/2x3 + 1/3x4 + 1/4x5 + .... + 1/98x99 + 1/99x100
= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/98 - 1/99 + 1/99 - 1/100
= 1/2 - 1/100
= 50/100 - 1/100
= 49/100
